Page 97 - 6099

Now we solve the equation for T(t).
0
T = −κλ n T
T = c e −κλ nt

The eigen-solutions of the partial differential equation that satisfy the homogeneous
boundary conditions are
r
2 p
u n (x, t) = sin λ n x e −κλ nt .
h

We seek a solution of the problem that is a linear combination of these eigen-solutions.

∞ r
X 2 p
u(x, t) = a n sin λ n x e −κλ nt
h
n=1
We apply the initial condition to find the coefficients in the expansion.

∞ r
X 2 p
u(x, 0) = a n sin λ n x = f(x)
h
n=1
r Z h
2 p
a n = sin λ n x f(x) dx
h 0

9.3 Time-Independent Sources and Boundary

Conditions

Consider the temperature in a one-dimensional rod of length h. The ends are held at
temperatures α and β, respectively, and the initial temperature is known at time t = 0.
Additionally, there is a heat source, s(x), that is independent of time. We find the
temperature by solving the problem,

u t = κu xx + s(x), u(0, t) = α, u(h, t) = β, u(x, 0) = f(x). (9.8)

Because of the source term, the equation is not separable, so we cannot directly apply
separation of variables. Furthermore, we have the added complication of inhomogeneous
boundary conditions. Instead of attacking this problem directly, we seek a transformation
that will yield a homogeneous equation and homogeneous boundary conditions.
Consider the equilibrium temperature, µ(x). It satisfies the problem,

s(x)
00
µ (x) = − = 0, µ(0) = α, µ(h) = β.
κ
The Green function for this problem is,

x < (x > − h)
G(x; ξ) = .
h

90